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bài 11
a) \(x^2-xy+x\\ =x\left(x-y+1\right)\)
b)
\(x^2-2xy-4+y^2\\ =\left(x^2-2xy+y^2\right)-4\\ =\left(x-y\right)^2-4\\ =\left(x-y-2\right)\left(x-y+2\right)\)
c)
\(x^3-x^2-16x+16\\ =x^2\left(x-1\right)-16\left(x-1\right)\\ =\left(x-1\right)\left(x-4\right)\left(x+4\right)\)
bài 12
\(2x\left(x-5\right)-x\left(3+2x\right)=26\)
\(2x^2-10x-3x-2x^2=26\)
\(-13x=26\\ x=-2\)
b)
\(2\left(x+5\right)-x^2-5x=0\\ 2\left(x+5\right)-x\left(x+5\right)=0\\ \left(x+5\right)\left(2-x\right)=0\\ \left[{}\begin{matrix}x+5=0\\2-x=0\end{matrix}\right.\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
a/\(\left(2x+1\right)\left(x+1\right)^2\left(2x+3\right)=18\)
\(\Leftrightarrow4x^4+16x^3+23x^2+14x-15=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x+5\right)\left(x^2+2x+3\right)=0\)
Tới đây thì đơn giản rồi b tự làm nhé
b/ \(3x^4-13x^3+16x^2-13x+3=0\)
\(\Leftrightarrow\left(x-3\right)\left(3x-1\right)\left(x^2-x+1\right)=0\)
Tới đây thì bạn làm tiếp nhé
c/ \(\left(x+3\right)^4+\left(x+5\right)^4=16\)
\(\Leftrightarrow2x^4+32x^3+204x^2+608x+690=0\)
\(\Leftrightarrow\left(x+3\right)\left(x+5\right)\left(x^2+8x+23\right)=0\)
Bạn làm tiếp nhé
a) \(x^3-16x=0\)
\(\Leftrightarrow x\left(x^2-16\right)=0\)
\(\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x-4=0\\x+4=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=4\\x=-4\end{array}\right.\)
b) \(\left(2x-3\right)^2=\left(x-5\right)^2\)
\(\Leftrightarrow\left(2x-3\right)^2-\left(x-5\right)^2=0\)
\(\Leftrightarrow\left(2x-3+x-5\right)\left(2x-3-x+5\right)=0\)
\(\Leftrightarrow\left(3x-8\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=\frac{8}{3}\end{array}\right.\)
c) \(x^2\left(x-1\right)-4x^2+8x-4=0\)
\(\Leftrightarrow x^2\left(x-1\right)-4\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=2\end{array}\right.\)
a,\((x+4)^2-(x+1)(x-1)=16\)
\(\Rightarrow x^2+8x+16-x^2+1=16\)
\(\Rightarrow 8x=-1\Rightarrow x=-\dfrac{1}{8}\)
b,\((2x-1)^2-(x+3)^2-5(x+7)(x-7)=0\)
\(\Rightarrow 4x^2-4x+1-(x^2+6x+9)-5(x^2-49)=0\)
\(\Rightarrow 4x^2-4x+1-x^2-6x-9-5x^2-245=0\)
\(\Rightarrow -x^2-10x-244=0\)
\(\Rightarrow -(x^2-10x+25)-219=0\)
\(\Rightarrow -(x-5)^2-219=0\)
\(\Rightarrow (x-5)^2+219=0\)
Mà \((x-5)^2+219>0\) suy ra PT vô nghiệm
a) \(25x^2-9=0\)
\(\Leftrightarrow\left(5x\right)^2-3^2=0\)
\(\Leftrightarrow\left(5x+3\right)\left(5x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}5x-3=0\\5x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{5}\\x=-\frac{3}{5}\end{cases}}\)
Vậy \(S=\left\{\frac{3}{5};\frac{-3}{5}\right\}\)
b) \(\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=16\)
\(\Leftrightarrow\left(x^2+8x+16\right)-\left(x^2-1\right)=16\)
\(\Leftrightarrow x^2+8x+16-x^2+1=16\)
\(\Leftrightarrow8x+17=16\)
\(\Leftrightarrow8x=-1\)
\(\Leftrightarrow x=-\frac{1}{8}\)
Vậy.........
c)\(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)
\(\Leftrightarrow\left(4x^2-4x+1\right)+\left(x^2+6x+9\right)-5\left(x^2-49\right)=0\)
\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5x^2+245=0\)
\(\Leftrightarrow2x=-255\)
\(\Leftrightarrow x=-127,5\)
Vậy.............
có j sai xót mong m.n bỏ qua☺
a) \(25x^2-9=0\)
<=> \(\left(5x\right)^2=9\)
<=> \(\left(5x\right)^2=3^2\)
<=> \(5x=3\)
<=> \(x=\frac{3}{5}\)
b) \(\left(x+4\right)^2-\left(x-1\right)\left(x+1\right)=16\)
<=> \(x^2+2.x.4+4^2-\left(x^2-1^2\right)=16\)
<=> \(x^2+8x+16-x^2+1=16\)
<=> \(\left(x^2-x^2\right)+8x+\left(16+1\right)=16\)
<=> \(8x+17=16\)
<=> \(8x=-1\)
<=> \(x=\frac{-1}{8}\)
c) \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)
<=> \(\left(2x\right)^2-2.2x.1+1^2+x^2+2.x.3+3^2-5\left(x^2-7^2\right)=0\)
<=> \(4x^2-4x+1+x^2+6x+9-5x^2+5.7^2=0\)
<=> \(\left(4x^2+x^2-5x^2\right)-\left(4x-6x\right)+\left(1+9+5.7^2\right)=0\)
<=> \(2x+245=0\)
<=> \(2x=-245\)
<=> \(x=\frac{-245}{2}\)
a) ... \(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x+2\right)=0\Leftrightarrow\hept{\begin{cases}x=1\\x=2\\x=-2\end{cases}}\)Vậy.....
b) ... \(\Leftrightarrow x^3\left(x-2\right)+10x\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3+10x\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x^2+10\right)=0\Leftrightarrow\hept{\begin{cases}x=0\\x=2\\x^2=-10\Rightarrow x\in\theta\end{cases}}\)(\(\theta\)là rỗng) Vậy.........
c) ... \(\Leftrightarrow2x-3=x+5\Leftrightarrow x=8\)Vậy.......
d) ... \(\Leftrightarrow x\left(x^2-16\right)=0\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\Leftrightarrow\hept{\begin{cases}x=0\\x=4\\x=-4\end{cases}}\)Vậy......
\(\frac{1}{3}x\left(x^2-16\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{3}x=0\\\left(x^2-16\right)=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2=16\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}\)
\(a.\frac{1}{3}x^3-\frac{16}{3}=0\)
\(\Rightarrow\frac{1}{3}x^3=0+\frac{16}{3}=\frac{16}{3}\)
\(\Rightarrow x^3=\frac{16}{3}:\frac{1}{3}=\frac{16}{9}\)
\(\Rightarrow x=\sqrt[3]{\frac{16}{9}}=1.21........\)